Question

Chứng minh

\(\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\) <1

Answer 1

\(A=\frac{3^2}{20.23}+\frac{3^2}{23.26}+...+\frac{3^2}{77.80}\)

\(\frac{A}{3}=\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\)

\(\frac{A}{3}=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\)

\(\frac{A}{3}=\frac{1}{20}-\frac{1}{80}\)

\(\frac{A}{3}=\frac{3}{80}\)

\(A=\frac{3}{80}.3=\frac{9}{80}< 1\)

Answer 2

Đặt A=3²/20.23+3²/23.26+^{....................}+3²/77.80

      A=3(3/20.23+3/23.26+.........+3/77.80)

     A=3(1/20-1/23+1/23-1/26+.+1/77-1/80)

     A=3(1/20-1/80)

    A=3.3/80

    A=9/80                       Mà A=9/80<1         =>A<1                   (đpcm)